Why does "" < {} evaluate to true?

Question

In JavaScript, why does:

"" < {} evaluate to true and

"" < [] evaluate to false?

Answer 1

Because < coerces its arguments. It prefers to compare numbers, but if the objects don't support a numeric primitive value, it does strings and you end up doing a lexical comparison. String({}) is "[object Object]", but String([]) is "". "" < "[object Object]" is true, but "" < "" is false.

All the gory details are in the spec (fair warning, the language is turgid to put it mildly).

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From your comment:

If it tries numbers first, consider that Number({}) is NaN and Number([]) is 0. Comparing NaN < 0 evaluates to false, and so does 0 < NaN. Why are these results ignored?

I put that poorly originally when I said originally that "...it tries numbers first..." (I've updated that). It doesn't. It just prefers numbers. Again, full details in soporific detail in the spec (various links from the above), but basically:

• The < operation does the abtract ToPrimitive operation on its operands.
• For objects, that invokes the internal [[DefaultValue]] method with the "hint" "Number".
• [[DefaultValue]] (hint = Number) invokes the valueOf method of the objects and, if that method returns a primitive, returns it; if the result wasn't a primitive, [[DefaultValue]] returns the result of toString instead. The valueOf method of objects (including arrays) returns the original object unchanged, so the result is [[DefaultValue]] returning the result of toString.
• The < operation sees that the operands are both strings and compares them lexically.

Whereas if the operands were primitive numbers, ToPrimitive would return the numbers unchanged, and the < would compare them mathematically; if they were Number instances (remember JavaScript has both primitive and object versions of numbers, strings, and booleans), Number#valueOf would be called by [[DefaultValue]], and Number#valueOf returns the primitive number value. And so < would compare them mathmatically.

Fun, eh?